2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313K. If the specific heat of air at constant volume is `0.718 k. j//kg K`, find the amount of heat absorbed in kj and kcal (`J=4.2 KJ// Kcal)`

Explanation

Mass of air , `m=2 kg`
Change in temperature, `Delta T=313-293=20K`
Specific heat at constant volune, `C_(v)=0.718K. J//kg-K`.
Heat mechanical equivalent, `J=4.2 KJ//K. cal. C_(v)=(1)/(m) (dQ)/(dT) rArr dQ =m C_(v)dT` or ` dQ =(mC_(v) dT)/(J) cal`.
therefore Heat energy absorbed, `Q=2xx0.718xx10^(3)xx20=28.72 KJ =6.838 K` calories.